Let $T_1, T_2\in\mathcal{L}(H)$ be two bounded operators on the infinite dimensional Hilbert space H.
My question is: If $R(T_1)=R(T_2)$ , then what could we say about $T_1,T_2$ ? Does there exist any relation between $T_1$ and $T_2$ ?
I know that when H is finite dimensional and $R(T_1)=R(T_2),$ then $T_1=T_2 S$ for some invertible operator $S \in \mathcal{L}(H)$ and the converse is also true. Will this result hold for infinite dimensoinal Hilbert space. Provide some opinions.
Thanks.
Pretty much the same trick seems to work for infinite dimensional Hilbert space when the o.n.b for $N(T_1)$ and $N(T_2)$ are of same cardinality.
Consider two orthornomal basis $\wedge_1$ and $\wedge_2$ for $N(T_1)$ and $N(T_2)$ respectively. Extend these bases to orthonormal basis $I_1$ and $I_2$ of $H$. Let $H_1=\overline{span\{I_1\setminus\wedge_1\}}$ and $H_2=\overline{span\{I_2\setminus \wedge_2\}}$. Notice that $T_1$ and $T_2$ restricted to $H_1$ and $H_2$ are injective map onto $R(T_1)=R(T_2)$ respectively (and hence isomorphism by open mapping theorem). Thus we can define a map $S:H_1\to H_2$ as $S(x)={T_2}_{|H_2}^{-1}{T_1}_{|H_1}(x)$. It is easy to check that this map is well defined, bijective, bounded and hence an isomorphism. Extend this isomorphism naturally to whole $H$ and then once can easily see that $T_1=T_2S$.
Infact $T_1=T_2S$ for some isomorphism $S$ forces that cardinality of o.n.b for $N(T_1)$ and $N(T_2)$ are same.